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Electric Circuit Theory
Nam Ki Min
010-9419-2320 nkmin@korea.ac.kr
Sinusoidal Steady-State Analysis
Chapter 9
Nam Ki Min
010-9419-2320 nkmin@korea.ac.kr
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 3 9.1 The Sinusoidal Source
Sinusoidal Source
Sinusoidal voltage source
β’ Produces a voltage that varies sinusoidally with time.
Sinusoidal current source
Sinusoidal source or sinusoidally time-varying excitation or sinusoid. β’ Produce a signal that has the form of the sine or cosine function.
β’ Produces a current that varies sinusoidally with time.
π£
π£
π
π
π£π
π ππ
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 4 9.1 The Sinusoidal Source
Alternating Circuit
A sinusoidal current is usually referred to as alternating current (ac).
Circuits driven by sinusoidal current or voltage sources are called ac circuits.
π 2
βπ 2
π 0
π π£(π‘)
AC Waveforms
Sinusoidal wave
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 5 9.1 The Sinusoidal Source
Why Sinusoids?
Nature itself is characteristically sinusoidal. We experience sinusoidal variation in the motion of a pendulum, the vibration of a string, the economic fluctuations of the stock market, and the natural response of underdamped second-order systemsβ
A sinusoidal signal is easy to generate and transmit.
It is the form of voltage generated throughout the world and supplied to homes, factories, laboratories, and so on.
It is the dominant form of signal in the communications and electric power industries.
Through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals.
A sinusoid is easy to handle mathematically.
π 2
βπ 2
π 0
π π£(π‘)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 6 9.1 The Sinusoidal Source
Sinusoidal Voltage
π£ π‘ = ππ sinππ‘
π£ π‘ = ππ cosππ‘
ππ cosππ‘ ππ sinππ‘
π£ π‘
ππ‘ π = 90Β°
π£ π‘ = ππ sin(ππ‘ Β± 90Β°) = Β±ππ cosππ‘
π£ π‘ = ππ cos(ππ‘ Β± 90Β°) = βππ sinππ‘
ππ
A sinusoid can be expressed in either sine or cosine form.
When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes.
This is achieved by using the following trigonometric identities:
sin(π΄ Β± π΅) = sin π΄ cosπ΅ Β± cosπ΄ sin π΅
cos(π΄ Β± π΅) = cosπ΄ cosπ΅ β sinπ΄ sinπ΅
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 7 9.1 The Sinusoidal Source
Sinusoidal Voltage
π£ π‘ = ππ cos(ππ‘ + π)
β’ ππ: amplitude
β’ π βΆ angular frequency in radians/s
β’ ππ‘: argument
β’ π βΆ period in second
π =2π
π
π£ π‘ + π = ππ cosπ π‘ + π = ππcos π π‘ +2π
π
= ππ cosππ‘
= ππ cos(ππ‘ + 2π)
= π£(π‘)
β’ π βΆ frequency in Hz
π =1
π π = 2ππ
A more general expression:
β’ π: phase angle
(9.1)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 8 9.1 The Sinusoidal Source
Converting Sine to Cosine
ππ cosππ‘
ππ sinππ‘
π£ π‘
ππ‘ π = 90Β°
ππ
π£ π‘ = ππ sin(ππ‘ + 90Β°) = ππ cosππ‘
π£ π‘ = ππ cos(ππ‘ β 90Β°) = ππ sinππ‘
π£ π‘ = ππ cos(ππ‘ Β± 180Β°) = βππ cosππ‘
π£ π‘ = ππ sin(ππ‘ Β± 180Β°) = βππ sinππ‘
π£ π‘ = ππ cos(5π‘ + 10Β°) = ππ sin(5t + 90Β° + 10Β°)
= ππ sin(5t + 100Β°)
π£(π‘) = ππ sin(5t β 260Β°)
= ππ sin(5t β 180Β° β 80Β°)
= βππ sin(5t + 90 β 170Β°)
= βππ cos(5t β 180 + 10Β°)
= ππ cos(5t + 10Β°)
= βππ sin(5t β 80Β°)
= βππ cos(5t β 170Β°)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 9 9.1 The Sinusoidal Source
Phase Relation of a Sinusoidal Wave
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 10 9.1 The Sinusoidal Source
Phase Shift (difference)
The two waves (A versus B) are of the same amplitude and frequency, but they are out of step with each other.
In technical terms, this is called a phase shift (difference).
ππ΄
ππ΅ π = 0
π£ π‘ = ππ sin(ππ‘ + ππ΄)
π£ π‘ = ππ sin(ππ‘ + ππ΅)
π£ π‘ = ππ cos(ππ‘ + ππ΄)
π£ π‘ = ππ cos(ππ‘ + ππ΅)
π
π ππ΄ ππ΅
π£ π‘ π£ π‘
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 11 9.1 The Sinusoidal Source
Phase Shift (difference)
Examples of phase shifts
The sinusoids are said to be in phase.
The sinusoids are said to be out of phase.
B βlagsβ A
A βlagsβ B
Leading
Lagging
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 12 9.2 The Sinusoidal Response
Sinusoidal Response
A sinusoidal forcing function produces both a natural (or transient) response and a forced (or steady-state) response, much like the step function, which we studied in Chapters 7 and 8.
The natural response of a circuit is dictated by the nature of the circuit, while the steady-state response always has a form similar to the forcing function.
However, the natural response dies out with time so that only the steady-state response remains after a long time.
When the natural response has become negligibly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state.
It is this sinusoidal steady-state response that is of main interest to us in this chapter.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 13
ππ
ππ‘+π
πΏπ =
πππΏcos(ππ‘ + π)
9.2 The Sinusoidal Response
βΊ Example β
For the circuit in Fi.9.5, find current
π£π (π‘) = ππ cos(ππ‘ + π)
πΏππ
ππ‘+ π π = ππ cos(ππ‘ + π)
π(π‘) =π
πΏ
π(π‘) =πππΏcos(ππ‘ + π)
π π‘ = exp π
πΏππ‘ = π
π πΏπ‘
(9.7)
(9.8)
(3)
(1)
(2)
(4)
π π₯ = exp π(π₯)ππ₯
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 14
π π‘ =1
ππ πΏπ‘ π
π πΏπ‘ πππΏcos(ππ‘ + π)ππ‘ +
πΎ
ππ πΏπ‘
9.2 The Sinusoidal Response
βΊ Example β
π π‘ = ππ πΏπ‘
π(π‘) =πππΏcos(ππ‘ + π) =
1
ππ πΏπ‘
πππΏ π
π πΏπ‘ cosππ‘ cosπ β sinππ‘ sinπ ππ‘ +
πΎ
ππ πΏπ‘
=1
ππ πΏπ‘
πππΏ
cosπ ππ πΏπ‘ cosππ‘ ππ‘ β sinπ π
π πΏπ‘ sinππ‘ ππ‘ +
πΎ
ππ πΏπ‘
cos(ππ‘ + π) = cosππ‘ cosπ β sinππ‘ sinπ
exp ππ₯ cos ππ₯ ππ₯ =πππ₯
π2 + π2(π cos ππ₯ + π sin ππ₯)
exp ππ₯ sin ππ₯ ππ₯ =πππ₯
π2 + π2(π sin ππ₯ β π cos ππ₯)
(5)
(6)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 15 9.4 The Passive Circuit Elements in the Frequency Domain
=1
ππ πΏπ‘
πππΏ
cosπ ππ πΏπ‘ cosππ‘ ππ‘ β sinπ π
π πΏπ‘ sinππ‘ ππ‘ +
πΎ
ππ πΏπ‘
exp ππ₯ cos ππ₯ ππ₯ =πππ₯
π2 + π2(π cos ππ₯ + π sin ππ₯)
exp ππ₯ sin ππ₯ ππ₯ =πππ₯
π2 + π2(π sin ππ₯ β π cos ππ₯)
βΊ Example β
=1
ππ πΏπ‘
πππΏ
cosππππ₯
π2 + π2 (π cos ππ‘ + π sin ππ‘) β sinπ
πππ₯
π2 + π2 (π sin ππ‘ β π cos ππ‘) +
πΎ
ππ πΏπ‘
π cos ππ₯ + π sin ππ₯ = π2 + π2π
π2 + π2cos ππ₯ +
π
π2 + π2sin ππ₯ = π2 + π2 cosπ cos ππ₯ + sin π sin ππ₯
π =π
πΏ
π = π πππ₯ = π
π πΏπ‘
=πππΏ
cosπ1
π2 + π2 (π cos ππ‘ + π sin ππ‘) β sinπ
1
π2 + π2 (π sin ππ‘ β π cos ππ‘) +
πΎ
ππ πΏπ‘
= π2 + π2 cos(π β ππ₯)
π sin ππ₯ β π cos ππ₯ = = π2 + π2 sin(ππ₯ β π) π2 + π2 cos π sin ππ₯ β sin π cos ππ₯
(7)
(8)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 16 9.2 The Sinusoidal Response
βΊ Example β
π cos ππ₯ + π sin ππ₯ = π2 + π2π
π2 + π2cos ππ₯ +
π
π2 + π2sin ππ₯ = π2 + π2 cosπ cos ππ₯ + sin π sin ππ₯
= π2 + π2 cos(π β ππ₯)
π sin ππ₯ β π cos ππ₯ = = π2 + π2 sin(ππ₯ β π) π2 + π2 cos π sin ππ₯ β sin π cos ππ₯
=πππΏ
cosπ1
π2 + π2π2 + π2 cos(π β ππ‘ β sinπ
1
π2 + π2π2 + π2 sin(ππ‘ β π) +
πΎ
ππ πΏπ‘
=πππΏ
1
π2 + π2cosπ cos(π β ππ‘) β sinπ sin(ππ‘ β π) +
πΎ
ππ πΏπ‘
=πππΏ
1
π πΏ
2
+ π2
cosπ cos(ππ‘ β π) β sinπ sin(ππ‘ β π) +πΎ
ππ πΏπ‘
π =π
πΏ
π = π
=πππΏ
cosπ1
π2 + π2 (π cos ππ‘ + π sin ππ‘) β sinπ
1
π2 + π2 (π sin ππ‘ β π cos ππ‘) +
πΎ
ππ πΏπ‘
(9)
(10)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 17 9.2 The Sinusoidal Response
βΊ Example β
=πππΏ
1
π πΏ
2
+ π2
cosπ cos(ππ‘ β π) β sinπ sin(ππ‘ β π) +πΎ
ππ πΏπ‘
π(π‘) = ππ1
π 2 + (ππΏ)2cos(ππ‘ + π β π) +
πΎ
ππ πΏπ‘
π =π
πΏ
π = π
0 = ππ1
π 2 + (ππΏ)2cos(π β π) + πΎ πΎ = βππ
1
π 2 + (ππΏ)2cos(π β π)
π(π‘) =ππ
π 2 + ππΏ 2cos(ππ‘ + π β π) β
ππ
π 2 + ππΏ 2cos(π β π) πβ
π πΏπ‘
(10)
(11)
(12)
(13) (9.9)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 18 9.2 The Sinusoidal Response
π π‘ = βππ
π 2 + ππΏ 2cos(π β π) πβ
π πΏπ‘ +
ππ
π 2 + ππΏ 2cos(ππ‘ + π β π) (9.9)
Sinusoidal Response
Transient(Natural) Component
β’ The natural response of a circuit is
dictated by the nature of the circuit.
β’ The natural response dies out with time.
Steady-state Component
(Forced Response)
β’ The steady-state solution is a sinusoidal function.
β’ The frequency of the response signal is identical to
the frequency of the source signal.
β’ The maximum amplitude of the steady-state
response, in general, differs from the maximum
amplitude of the source.
β’ The phase angle of the response signal, in general,
differs from the phase angle of the source.
When the natural response has become
negligibly small compared with the steady-
state response, we say that the circuit is
operating at sinusoidal steady state.
It is this sinusoidal steady-state response
that is of main interest to us in this chapter.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 19 9.3 The Phasor
Definition
A phasor is a complex number that represents the amplitude and phase angle of a sinusoid.
π£ π‘ = ππ cos(ππ‘ + π) β π = πππππ = ππβ π
phasor representation
When a phasor is used to describe an AC quantity, the length of a phasor represents the amplitude of the wave while the angle of a phasor represents the phase angle of the wave relative to some other (reference) waveform.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 20
Complex Number
A complex number z can be written in rectangular form as
π§ = π₯ + ππ¦
π = β1
Representation of a complex number z in the complex plane
π₯: the real part of z
the imaginary part of z π¦:
The complex number z can also be written in polar or exponential form as
π§ = ππππ
π = π₯2 + π¦2
π = π‘ππβ1π¦
π₯
π§ = πβ π
The relationship between the rectangular form and the polar form is
π§ = π₯ + ππ¦ = π(cosπ + π sinπ) = πβ π
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 21
Complex Number
Addition and subtraction of complex numbers
π§1 = π₯1 + ππ¦1 = π1β π1
Multiplication and division
π§1π§2 = π1π2β π1 + π2
Reciprocal
Square Root
π§2 = π₯2 + ππ¦2 = π2β π2
π§1 + π§2 = (π₯1 + π₯2) + π(π¦1 + π¦2)
π§1 β π§2 = (π₯1 β π₯2) + π(π¦1 β π¦2)
π§1π§2=π1π2β π1 β π2
1
π§=
1
π₯ + ππ¦ =
1
π₯2 + π¦2β β π
1
π§=
1
πβ π =1
π β β π
π§ = πβ π/2
9.3 The Phasor
Representation of a complex number z in the complex plane
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 22
Complex Conjugate
Complex Conjugate
π§ = π₯ + ππ¦
π§β = π₯ β ππ¦ = πβ β π = ππβπ
βπ =1
π β
1
π=
1
1β 90Β°=1
1β β 90Β° = βπ
9.3 The Phasor
Representation of a complex number z in the complex plane
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 23
The Phasor Representation of the Sinusoid v(t)
The idea of phasor representation is based on Eulerβs identity. In general,
πΒ±ππ = cos π + π sin π (9.10)
We can regard cos π and sin π as the real and imaginary parts of πππ; we may write
cos π = Re{πππ}
sin π = Im{πππ}
(9.11)
(9.12)
We write the sinusoidal voltage function given by Eq.(9.1) in the form suggested by Eq.(9.11)
(9.1) π£ π‘ = ππ cos(ππ‘ + π)
= Re{ππππ ππ‘+π }
= Re{πππππππππ‘}
πππππ : a complex number that carries the amplitude and phase angle
of the given sinusoidal function.
We define the phasor representation or phasor transform of the given sinusoidal function as
π£ π‘ = Re{πππππ‘} π = πππππ (9.15)
(9.14)
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 24
The Phasor Representation of the Sinusoid v(t)
Phasor representation
π = πππππ (9.15)
π = ππβ π
π = ππ(cosπ + π sinπ) (9.16)
One way of looking at Eqs. (9.15) and (9.16) is to consider the plot of the πππππ‘ on the complex plane.
As time increases, the πππππ‘ rotates on a circle of radius ππ at an angular velocity Ο in the counterclockwise direction, as shown in Fig. (a). In other words, the entire complex plane is rotating at an angular velocity of Ο.
We may regard π£(π‘) as the projection of the πππππ‘ on the real axis, as shown in Fig.(b).
The value of the πππππ‘ at time t = 0 is the phasor π of the sinusoid π£(π‘). The πππππ‘ may be regarded as a rotating phasor.
Thus, whenever a sinusoid is expressed as a phasor, the term ππππ is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency Ο of the phasor; otherwise we can make serious mistakes.
π
πππππ‘
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 25
Phasor Transform : Summary
Eqs.(9.14) through (9.16) reveal that to get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we take out the time factor ππππ‘, and whatever is left is the phasor corresponding to the sinusoid.
By suppressing the time factor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows:
= Re{πππππππππ‘} π£ π‘ = ππ cos(ππ‘ + π)
= Re{πππππ‘}
π = πππππ
= ππβ π (1)
Time-domain representation Phasor or frequency-domain representation
ππ sin(ππ‘ + π) = ππ cos(ππ‘ + π β 90Β°)
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 26
Inverse Phasor Transform
Equation (1) states that to obtain the sinusoid corresponding to a given phasor V, multiply the phasor by the time factor ππππ‘ and take the real part.
π = 115β β 45Β°
π = ππβ π
π = 500 rad/s
= ππ cos(ππ‘ + π)
π£ π‘ = 115 cos(500π‘ β 45Β°)
π£(π‘) = Rm{πππππππππ‘} = πππ
ππ
π = π8πβπ20Β°
= (1β 90Β°)(8β β 20Β°)
= 8β 90Β° β 20Β°
= 8β 70°° π£ π‘ = 8 cos(ππ‘ + 70Β°)
π = π 5 β π12 = 12 + π5
= 122 + 52β π π = tanβ15
12= 22.62Β°
π£ π‘ = 13 cos(ππ‘ + 22.62Β°) = 13β 22.62Β°
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 27
Phasor Diagram
Since a phasor has magnitude and phase (βdirectionβ), it behaves as a vector and is printed in boldface.
For example, phasors π = ππβ π and π = πΌπβ β π are graphically represented in Figure.
Such a graphical representation of phasors is known as a phasor diagram.
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 28
Phasor Diagram
The relationship between two phasors at the same frequency remains constant as they rotate; hence the phase angle is constant.
Consequently, we can usually drop the reference to rotation in the phasor diagrams and study the relationship between phasors simply by plotting them as vectors having a common origin and separated by the appropriate angles.
Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency.
Phasor Diagram
π
π
π
π
π
π π
π£(π‘) = ππ cos(ππ‘ + π2)
π(π‘) = πΌπ cos(ππ‘ + π1)
π = π½π β π2
π = π°π β π1 π = ππ βππ
9.3 The Phasor
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 29 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for a Resistor>
Time Domain
β’ Current
π(π‘) = πΌπ cos(ππ‘ + π)
β’ Voltage
π£ π‘ = ππ = π πΌπ cos(ππ‘ + π)
Phasor form
π = πΌπβ π
π = π π = π πΌπβ π
Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.
This shows that the voltage-current relation for the resistor in the phasor domain continues to be Ohmβs law, as in the time domain.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 30 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for a Resistor>
Phasor Diagram
π = πΌπβ π
π = π π = π πΌπβ π
π = πΌπβ 0Β°
π = π π = π πΌπβ 0Β°
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 31 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for an Inductor>
Time Domain
π(π‘) = πΌπ cos(ππ‘ + π)
π£ π‘ = πΏππ
ππ‘= βππΏπΌπ sin(ππ‘ + π) π£
π
= βππΏπΌπ cos(ππ‘ + π β 90Β°)
= ππΏπΌπ cos(ππ‘ + π + 90Β°)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 32 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for an Inductor>
Phasor form
π = πΌπβ π
π = ππΏπΌπβ π + 90Β°
π£(π‘) = ππΏπΌπ cos(ππ‘ + π + 90Β°)
= (1β 90Β°)(ππΏπΌπβ π)
π = 1β 90Β° = πππΏπΌπβ π
= πππΏπ
Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 33 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for an Inductor>
ππ
ππ‘ = ππΌπ cos(ππ‘ + π + 90Β°)
= Re{ππΌπ ππππ‘πππππ90Β° }
= Re{πππΌπ ππππ‘πππ }
= Re{πππ ππππ‘ }
ππ
ππ‘
π = πΌπβ π π(π‘) = πΌπ cos(ππ‘ + π)
πππ
Differentiating a sinusoid is equivalent to
multiplying its corresponding phasor by ππ.
π£ π‘ = πΏππ
ππ‘= πΏπππ = πππΏπ
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 34
Phasor Diagram
9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for an Inductor>
π = πΌπβ 0Β°
= πππΏπ
Voltage leads current by 90Β°in an inductor
Current lags voltage by 90Β° in an inductor
π = ππΏπΌπβ 90Β°
π = πΌπβ π
π = ππΏπΌπβ π + 90Β°
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 35 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for a Capacitor>
Time Domain
π£(π‘) = ππ cos(ππ‘ + π)
π π‘ = πΆππ£
ππ‘= βππΆππ sin(ππ‘ + π)
= βππΆππ cos(ππ‘ + π β 90Β°)
= ππΆππ cos(ππ‘ + π + 90Β°)
π£
π
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 36 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for a Capacitor>
Phasor form
π = ππβ π
π = ππΆππβ π + 90Β°
π(π‘) = ππΆππ cos(ππ‘ + π + 90Β°)
= (1β 90Β°)(ππΆππβ π)
π = 1β 90Β° = πππΆππβ π
= πππΆπ
π =1
πππΆπ
Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 37 9.4 The Passive Circuit Elements in the Frequency Domain
< The V- I Relationship for a Capacitor>
Phasor Diagram
π = ππβ π
π = ππΆππβ π + 90Β°
π = ππβ 0Β°
π = ππΆππβ 90Β°
= πππΆπ
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 38 9.4 The Passive Circuit Elements in the Frequency Domain
Summary
Summary of voltage-current relationships
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 39 9.4 The Passive Circuit Elements in the Frequency Domain
< Impedance, Reactance, and Admittance>
Impedance
We obtained the voltage-current relations for the three passive elements as
π = π π
π =1
πππΆπ
π = πππΏπ
π
π= π
π
π= πππΏ
π
π =
1
πππΆ
From these three expressions, we obtain Ohmβs law in phasor form for any type of element as
π =π
π π = ππ or
Z is a frequency-dependent quantity known as impedance, measured in ohms.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 40 9.4 The Passive Circuit Elements in the Frequency Domain
< Impedance, Reactance, and Admittance>
Impedance
The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms.
The impedance represents the opposition which the circuit exhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity.
The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.1 summarizes their impedances.
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 41 9.4 The Passive Circuit Elements in the Frequency Domain
< Impedance, Reactance, and Admittance>
Resistance and Reactance
As a complex quantity, the impedance may be expressed in rectangular form as
π =π
π= π + ππ
π : Resistance
π : Reactance
Inductive and capacitive reactance
π = π + ππ
π<0 :
π>0 :
π = π β ππ
Inductive reactance
Capacitive reactance
π = π
π = πππΏ
π =1
πππΆ= βπ
1
ππΆ
Resistor :
Inductor :
Capacitor:
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 42 9.4 The Passive Circuit Elements in the Frequency Domain
< Impedance, Reactance, and Admittance>
Impedance in Polar Form
π = π + ππ
π = πβ π π = π 2 + π2 π = tanβ1
π
π
π
π
ππ
π
π = π§ cos π
π = π sin π
Admittance
It is sometimes convenient to work with the reciprocal of impedance, known as admittance.
The admittance Y is the reciprocal of impedance, measured in siemens (S).
π =1
π=π
π
As a complex quantity, we may write Y as
π = πΊ + ππ΅
πΊ : Conductance
π΅ : Susceptance
π =1
π + ππ =π β ππ
π 2 + π2
πΊ =π
π 2 + π2 π΅ = β
π
π 2 + π2
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 43 9.4 The Passive Circuit Elements in the Frequency Domain
< Impedance, Reactance, and Admittance>
π = πΊ + ππ΅
πΊ : Conductance
π΅ : Susceptance
π =1
π + ππ =π β ππ
π 2 + π2
πΊ =π
π 2 + π2 π΅ = β
π
π 2 + π2
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 44 9.4 The Passive Circuit Elements in the Frequency Domain
βΊ Example β
(a) Convert the circuit to the frequency domain.
(b) Find an admittance for resistor and capacitor
πs = 10 β 0Β° V
πR = 5 Ξ©
πc = βπ1
ππΆ= βπ
1
4 Γ 0.1= βπ2.5 Ξ©
πR =1
5= 0.2 S
πc = πππΆ = π4 Γ 0.1 = π0.4 S
Voltage phasor
Impedance
Admittance
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 45 9.5 Kirchhoffβs Laws in the Frequency Domain
KVL in the frequency Domain
For KVL, let π£1, π£2, β― , π£π be the voltages around a closed loop. Then
π£1 + π£2 +β―+ π£π = 0
In the sinusoidal steady state, each voltage may be written in cosine form, so that Eq. (9.36) becomes
(9.36)
ππ1 cos(ππ‘ + π1) + ππ2 cos(ππ‘ + π2) + β―+ πππ cos(ππ‘ + ππ) = 0 (9.37)
We now Eulerβs identity to write Eq.(9.37) as
Re ππ1πππ1ππππ‘} + Re{ππ2π
ππ2ππππ‘} + β―+ Re{πππππππππππ‘} = 0 (9.38)
or
Re (ππ1πππ1 + ππ2π
ππ2 +β―+ πππππππ)ππππ‘} = 0
If we let ππ = πππππππ, then
Re (π1 + ππ +β―+ ππ)ππππ‘} = 0
Since ππππ‘ β 0,
π1 + ππ +β―+ ππ = 0
(9.39)
(9.40)
(9.41)
indicating that Kirchhoffβs voltage law holds for phasors.
+ π£1 β + π£2 β
+ π£π β
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 46 9.5 Kirchhoffβs Laws in the Frequency Domain
KCL in the frequency Domain
By following a similar procedure, we can show that Kirchhoffβs current law holds for phasors. If we let π1, π2, β― , ππ , in be the current leaving or entering a closed surface in a network at time t, then
π1 + π2 +β―+ ππ = 0
If π1, ππ, β― , ππ, are the phasor forms of the sinusoids π1, π2, β― , ππ , then
(9.42)
(9.43) π1 + ππ +β―+ ππ = 0
which is Kirchhoffβs current law in the frequency domain.
βΊ Example β
Find π(π‘) in the circuit
πs = 10 β 0Β° V
KVL :
πR = 5 Ξ© πc = βπ2.5 Ξ©
πs = πR + πC = ππR + ππC = 5π + βj2.5 π
π =10β 0Β°
5 β π2.5=10(5 + π2.5)
52 + 2.52= 1.6 + π0.8 = 1.789β 26.57Β° A
π1
π2
ππ
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 47 9.6 Series, Parallel, and Delta-to-Wye Simplifications
Series Connection
showing that the total or equivalent impedance of series-connected impedances is the sum of the individual impedances.
This is similar to the series connection of resistances.
πab = ππ + ππ +β―+ ππ
= ππ1 + πππ +β―+ πππ§
= π(π1 + ππ +β―+ ππ§)
πab =πabπ= π1 + ππ +β―+ ππ§
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 48 9.4 The Passive Circuit Elements in the Frequency Domain
βΊ Example β
For the circuit in Fi.9.5, find (a) current, (b) impedance, and (c) reactance.
π£π (π‘) = ππ cos(ππ‘ + π)
β’ Phasor form
π = π + πππΏ = πβ π
ππ = ππβ π
π =ππ π=
ππβ π
π 2 + (ππΏ)2β π
π = π 2 + (ππΏ)2
π = tanβ1ππΏ
π
Voltage :
Impedance :
π
π
πππΏ
π
π = π§ cos π
π = π sin π Current : =
ππβ π β π
π 2 + (ππΏ)2
π π‘ =ππ
π 2 + ππΏ 2cos(ππ‘ + π β π)
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 49 9.6 Series, Parallel, and Delta-to-Wye Simplifications
Parallel Connection
This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.
π = ππ + ππ +β―+ ππ
= π1
ππ+1
π2+β―+
1
πn =
π
πππ
1
πππ=1
ππ+1
π2+β―+
1
πn
πππ = π1 + π2 +β―+ ππ
CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 nkmin@korea.ac.kr
Chapter 9 Sinusoidal Steady-State Analysis 50 9.6 Series, Parallel, and Delta-to-Wye Simplifications
Delta-to-Wye Transformation
Z1 =ππππ
ππ + ππ + ππ , (9.51)
Z2 =ππππ
ππ + ππ + ππ , (9.52)
Z3 =ππππ
ππ + ππ + ππ , (9.53)
Za =π1π2+ π2π3+ π3π1
π1 , (9.54)
Zb =π1π2+ π2π3+ π3π1
π2 , 9.55
Zc =π1π2+ π2π3+ π3π1
π3 , (9.56)
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